这是一份anu西澳大学MATH3511/MATH6111的成功案例

To be more specific, suppose that
$$
F(h)=a_{0}+a_{1} h^{p}+\mathcal{O}\left(h^{r}\right)
$$
as $h \rightarrow 0$ for some $p$ and $r$, with $r>p$. We assume that we know the values of $p$ and $r$, but not $a_{0}$ or $a_{1}$. Indeed, $F(0)=a_{0}$ is the quantity we seek. Suppose that we have computed $F$ for two stepsizes, say, $h$ and $q h$ for some $q>1$. Then we have
$$
F(h)=a_{0}+a_{1} h^{p}+\mathcal{O}\left(h^{r}\right)
$$
and
$$
F(q h)=a_{0}+a_{1}(q h)^{p}+\mathcal{O}\left(h^{r}\right) .
$$
This system of two linear equations in the two unknowns $a_{0}$ and $a_{1}$ is easily solved to obtain
$$
a_{0}=F(h)+\frac{F(h)-F(q h)}{q^{p}-1}+\mathcal{O}\left(h^{r}\right) .
$$
Thus, the accuracy of the improved value, $a_{0}$, is $\mathcal{O}\left(h^{r}\right)$ rather than only $\mathcal{O}\left(h^{p}\right)$.

MATH3511/MATH6111 COURSE NOTES ：

For example, applying Euler’s method to this equation using a fixed stepsize $h$, we have
$$
y_{k+1}=y_{k}+\lambda y_{k} h=(1+\lambda h) y_{k},
$$
which means that
$$
y_{k}=(1+\lambda h)^{k} y_{0} .
$$
Provided $\lambda<0$, the exact solution decays to zero as $t$ increases, as will the computed solution if $|1+\lambda h|<1$. This result agrees with our earlier stability analysis because $J=\lambda$ for this ODE. We also note that the growth factor $1+\lambda h$ agrees with the series expansion
$$
e^{\lambda h}=1+\lambda h+\frac{(\lambda h)^{2}}{2}+\frac{(\lambda h)^{3}}{6}+\cdots
$$
through terms of first order in $h$, and hence Euler’s method is first-order accurate. Especially for more complicated numerical methods, a linear ODE is easier to work with than a general ODE, and it produces essentially the same stability result if we equate $\lambda$ with the Jacobian $J$ at a given point. An important caveat, however, is that $\lambda$ is constant, whereas the Jacobian $J$ varies for a nonlinear equation, and hence the stability can potentially change.